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2 years ago

AbbyWinterHeart, Cookiemookielookie I NEED HELP!!!

Physics

2 answers:

pochemuha2 years ago

4 0

Answer:

the third option down is correct

PtichkaEL [24]2 years ago

3 0

I think it’s the third one. Not sure tho

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A storm 10.0 km in diameter has wind speeds of 10.0 m/s. What is its angular velocity in radians per second

Aloiza [94]

Answer:

The value is $w= 2*10^{-3} \ rad/s$

Explanation:

From the question we are told that

The diameter of the storm is $d = 10.0 \ km = 10 000 \ m$

The wind speed is $v = 10.0 \ m/s$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{10000}{2}$

=> $r = 5000 \ m$

Generally the angular velocity is mathematically represented as

$w= \frac{v}{r}$

=> $w= \frac{10}{5000}$

=> $w= 2*10^{-3} \ rad/s$

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2 years ago

Will give BRAINEST to the person who answers right

Nookie1986 [14]

The answer of this question is B. 22x + 20

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3 years ago

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A 50kg skater at rest on a frictionless rink throws a 2kg ball, giving the ball a velocity of 10m/s. Which statement describes t

Andreas93 [3]

0.4m/s in the opposite direction of the ball.

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2 years ago

What demonstrates the flow of energy and materials through

Tcecarenko [31]

A. Food chain

A food web also shows the flow of energy and materials through an ecosystem but in a complex web, not a single chain. A biogeochemical cycle does not deal with the flow of energy/materials through an ecosystem at all.

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10 months ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

Part A: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

Part B: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$